Answer
$ID_1=ID_2 + 1$.
$ID_1$ is $1$ more than $ID_2$.
Work Step by Step
$ID=\frac{\log(2A/W)}{\log 2},$ Whereas $W$=width of the target and, $A$=distance to the center of the target.
$A=100mm, W_1=5mm$.
$ID_1=\frac{\log(2(100)/5)}{\log 2},$
$ID_1=\frac{\log(200/5)}{\log 2},$
$ID_1=\frac{\log40}{\log 2}=\frac{1.6021}{0.3010}=5.322,$
$A=100mm, W_2=10mm$.
$ID_2=\frac{\log(2(100)/10)}{\log 2},$
$ID_2=\frac{\log(200/10)}{\log 2},$
$ID_2=\frac{\log20}{\log 2}=\frac{1.3010}{0.3010}=4.322,$
Therefore, $ID_1=ID_2 + 1$.
$ID_1$ is $1$ more than $ID_2$.
